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You need 20 bits of address bus to address 1 Mb of memory.
15 bits
You need 12 address lines to access 4K of memory. 212 = 4096. log2 (4096) = 12.
That depends on the memory architecture of the system.if the memory chips are byte wide and not used to create a multibyte bus, 11 address bits are needed.if the memory chips are 32 bits wide, 9 address bits are needed (with the CPU internally selecting which of the 4 bytes it will use).it the memory chips are 64 bits wide, 8 address bits are needed (with the CPU internally selecting which of the 8 bytes it will use.if the memory chips are 4 bits wide, 12 address bits will be needed and the CPU must perform 2 memory cycles per byte that it needs. (yes, I have seen a computer that worked this way!)etc.
4K (4096) of addressable space is defined by 12 bits of address space, because 212 = 4096.
When the low order bits of the address are used to select the memory bank it is interleaved.
4 pages -> 2^2 bits 1024 bytes -> 2^10 bits 64 frames -> 2^6 bits Therefore: Logical memory = 2+10=12 bits Physical memory = 10 +6 =16 bits
8 bits
It takes 23 address lines to address 8 mb of memory.
16384
It requires 30 address bits to address 1GB of RAM.230 = 1,073,741,824
8086 has 20 address lines. Therefore it can address 220 bits or 1,048,576 bits of memory, or roughly 1 MB (mega byte).